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Five cards are drawn from a shuffled deck with $52$ cards. Find the probability that

a) four cards are aces

b) four cards are aces and the other is a king

c) three cards are tens and two are jacks

d) at least one card is an ace

My attempt:

a) $\left(13*12*\binom{4}{4}*\binom{4}{1}\right)/\binom{52}{5}$

b) same as (a)?

c) $\left(13 * 12 * \binom{4}{3} * \binom{4}{2}\right)/\binom{52}{5}$

d) $\left(13 * \binom{4}{1}\right)/\binom{52}{5}$

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4 Answers 4

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a) There are $ \binom{52}{5} = 2,598,960 $ ways of choosing 5 cards. There are $\binom{4}{4} = 1 $ way to select the 4 aces. So there are $\binom{48}{1}= 48 $ ways to select the remaining card. Thus there are a total of 48 ways to select 5 cards such that 4 of them are aces, and the probability is: $\frac{48}{2,598,960} = \frac{1}{54,145}$.

b) There are $\binom{4}{4} = 1 $ way to choose the 4 aces, and there are $\binom{4}{1} = 4 $ ways to choose a king. So there are $1\times4 = 4 $ ways to choose 5 cards such that 4 are aces and the other is a king card. The probability is: $\frac{4}{2,598,960} = \frac{1}{649,740} $.

c) There are $\binom{4}{3} = 4 $ ways to choose 3 ten cards, and there are $\binom{4}{2} = 6 $ ways of choosing 2 jacks. So there are $ 4\times 6 = 24 $ ways to choose 5 cards such that 3 are ten and 2 are jacks. The probability for this case is: $\frac{24}{2,598,960} = \frac{1}{108,290} $.

d) The probability of 5 non-ace cards is: $\frac{\binom{48}{5}}{\binom{52}{5}} = \frac{1,712,304}{2,598,960} = 0.6588 $, so the probability of getting 5 card at least one ace is: $1 - 0.6588 = 0.34$.

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  • $\begingroup$ Could you guys explain why the guy below says multiply 13 and 12 into the numerator? $\endgroup$
    – rojas777
    Commented Mar 13, 2014 at 22:49
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There are $\binom{52}{5}$ equally likely ways to choose $5$ cards. For solving all but the last problem, we count the number of "favourables" and divide by $\binom{52}{5}$.

a) There are $\binom{48}{1}$ $4$-Ace hands, for we have freedom only in choosing the non-Ace. Or, if you prefer, there are $\binom{12}{1}\binom{4}{1}$ such hands.

b) We can choose which King it is in $\binom{4}{1}$ ways.

c), The tens can be chosen in $\binom{4}{3}$ ways, and for each way the Jacks can be chosen in $\binom{4}{2}$ ways, for a total of $\binom{4}{3}\binom{4}{2}$.

d) Find first the probability of no Aces. We can choose a no Ace hand in $\binom{48}{5}$ ways. Calculate the probability of no Aces, and subtract this probability from $1$.

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  • $\begingroup$ Do you know why user235235 says multiply 13 and 12 into the numerator? $\endgroup$
    – rojas777
    Commented Mar 14, 2014 at 16:37
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    $\begingroup$ The reason is a misreading of the question. For the probability a a full house, we indeed multiply by $(13)(12)$. However, Question c) asks for the probability of a specific type of full house, three $10$'s and two Jacks. $\endgroup$ Commented Mar 14, 2014 at 16:41
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I'm not sure the other people here are right. C is basically asking for a full house, isn't it?

According to Wiki, the probability is found by $$\frac{13 \times 12 \times \binom {4}{3} \times \binom 42} {\binom{52}5}$$

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    $\begingroup$ The probability you've cited here is the probability of drawing ANY full house. There are 156 different full houses possible (13 x 12). However, the original question requires a specific full house (tens full of Jacks), so the probability of that is much smaller. $\endgroup$
    – BBO555
    Commented Sep 19, 2018 at 9:36
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deepsea gave a complete and clear answer.

I'd just add that you could see straight away that the answers for (b) and (a) cannot be the same, because the requirement in (b) is so much more specific. Many hands that satisfy (a) do not satisfy the conditions for (b).

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